MHB Dale Simpson's question at Yahoo Answers (One-to-one matrix)

[Fernando Revilla]

Here is the question:

recall a matrix A is one-to-one if for vectors x and y, Ax = Ay implies that x = y. Suppose A is a
matrix with columns v1, v2, . . . vk. Prove if any linear combination of the form
a1*v1 + a2*v2 + : : : + ak*vk = 0;
has only the trivial solution, then A is one-to-one.

Here is a link to the question:

Prove a matrix is one to one? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Dale,

Suppose that $A$ (order $m\times k$) is not one-to-one, then there exist vectors $x=(x_1,\ldots,x_k)^t$ and $y=(y_1,\ldots,y_k)^t$ such that $Ax=Ay$ with $x\neq y$. Equivalently, $A(x-y)=0$ with $x\neq y$. We have $x_i-y_i\neq 0$ for some $i$. Then, $$\begin{aligned}&A\begin{bmatrix}x_1-y_1\\ \vdots\\{x_i-y_i}\\ \vdots\\x_k-y_k\end{bmatrix}=\begin{bmatrix}{v_1}&{\ldots}&{v_k}\end{bmatrix}\begin{bmatrix}x_1-y_1\\ \vdots\\{x_i-y_i}\\ \vdots\\x_k-y_k\end{bmatrix}\\&=(x_1-y_1)v_1+\ldots+(x_i-y_i)v_i+\ldots+(x_k-y_k)v_k=0\end{aligned}$$ This implies that not all linear combination $a_1v_1+\ldots+a_kv_k=0$ has only the trivial solution (contradiction).